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I don't completely understand the distributional character of a quantum field because I never see them "smeared" in basic textbooks. As I understand it, if $\chi : \mathcal{F} \rightarrow \mathbb{K}$ such that

$$ \chi[f] = \int\limits_{-\infty}^{\infty} f(x) \chi(x) \text{d} x,$$

then $\chi$ is said to be a distribution smeared with $f \in \mathcal{F}$; where $\mathcal{F}$ is the space of scalar functions and $\mathbb{K}$ is a number field. Now, let's take a free Klein-Gordon quantum field as an example. This field is expressed as

$$\phi(\textbf{x}, t) = \frac{1}{(2 \pi)^3} \int \frac{1}{\sqrt{2 E_{\textbf{p}}}} \Big(a_{\textbf{p}} (t) e^{-i p_\mu x^\mu} + a^\dagger_{\textbf{p}} (t) e^{i p_\mu x^\mu} \Big) \text{d}^3 p.$$

What confuses me is that, unless I'm wrong, this field is an operator valued distribution. If that's the case, shouldn't $\phi$ be smeared with a scalar function in order for it to be well defined? Explicitly, shouldn't it be written something like this

$$\phi[f] = \int f(x) \phi (x) \text{d} x,$$

such that, for $|\psi \rangle$ that belongs in a symmetric Fock space

$$\phi[f] |\psi \rangle = \Bigg( \int f(x) \phi (x) \text{d} x \Bigg) |\psi \rangle ?$$

If that's the case, this confuses me further, since the expression inside the parenthesis is a number. How could $|\psi \rangle$ be evaluated then?

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4 Answers 4

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Yes, the quantum fields must be smeared in order to become well-behaved (symmetric, densely defined) operators (in the Hilbert space of the theory). In mathematically-minded textbooks it is the standard of their initial presentation.

Actually, referring to coherent states they (the bosonic ones) could be viewed as quadratic forms without smearing, but this is a very special case, even if it takes into account some classical interpretation of these fields.

The point is that smeared quantum fields are not sufficient to define all physically interesting observables. As a consequence unsmeared quantum fields have its own physical relevance, though this notion is affected by several mathematical issues.

For instance, the stress-energy tensor or (self-)interaction terms, e.g. $\phi(x)^4$, in Lagrangians are linear combinations of products of unsmeared field operators. Here, the smearing occurs after taking the product.

This notion is mathematically not well defined and it needs some care to be used. At the end of the day, ultraviolet renormalization has its roots in this fact.

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  • $\begingroup$ How do we know then that QFT is a well "built" theory if unsmeared fields, although physically relevant, are affected by several mathematical issues as you say? Are we putting all our "faith" in experimental data or have there been attempts to make the theory more mathematically rigorous? And, as for what you said at the end, this has to do with normal ordering and Hadamard states, right? Sorry for the long comment. $\endgroup$
    – RMC777
    Nov 6, 2023 at 7:34
  • $\begingroup$ I wanted to ask, since we also have renormalization flow in lattice field theories which don't even have operator valued distributions, in what sense can renormalization be viewed as a result of products of distributions not being well defined? I mean how the two views of renormalization are connected. $\endgroup$
    – Ryder Rude
    Nov 6, 2023 at 8:02
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    $\begingroup$ @RMC QFT is the best we can do. E.g., QED is an astonishing product of it with many prediction confirmed, in spite of mathematical problems still affecting it. Yes normal ordering and Hadarmard states are part of the rigirous machinery which, however, did not fix basic issues. $\endgroup$ Nov 6, 2023 at 14:42
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    $\begingroup$ @RMC777 an important textbook is Local Quantum Physics by R Haag 2nd edition. A more recent book is this. link.springer.com/book/10.1007/978-3-319-21353-8 that collects several reviews (chapter 5 is mine with I. Khavkine) $\endgroup$ Nov 6, 2023 at 16:04
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    $\begingroup$ @RMC777: We don't actually have a unified and well-defined quantum theory that unites with special relativity. We have, instead, 3 separate well-defined theories of limited but overlapping range: free-field QFT, i.e. non-interacting; perturbative scattering QFT, which deals with momentary interactions between particles; and lattice QFT, which deals with continuous interaction but is not strictly relativistic (i.e. is not Lorentz-invariant), rather only approximately so. What we don't have is an exact, unified, interacting, fully-SR-compliant quantum mechanics. $\endgroup$ Nov 6, 2023 at 20:35
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The main point of quantum field theory is to study dynamics. The dynamics is given by interactions among the different fields. These interactions are point interactions. The way that these point interactions are modeled in the theory is with the aid of field operators that annihilate and create the different fields. To be able to represent point interactions, these field operators need to be defined at points and not be smeared out. Eventually, the interaction points are integrated over all space, which I guess takes over the role of the smearing process.

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  • $\begingroup$ So, if I understood you correctly, the smearing process occurs $\textit{after}$ the corresponding Fock state is evaluated on the field? $\endgroup$
    – RMC777
    Nov 6, 2023 at 4:00
  • $\begingroup$ Just to make sure I understand what you mean, is the Fock state you are referring to the assumed input state? Or what do you mean by a Fock state being "evaluated" on a field? $\endgroup$ Nov 6, 2023 at 4:04
  • $\begingroup$ Yes, I do mean the input state. Explicitly, I mean this $\phi(\textbf{x}, t) |\psi \rangle$. $\endgroup$
    – RMC777
    Nov 6, 2023 at 4:07
  • $\begingroup$ In QFT, there would be more stuff going on, leading to the smearing. If you just write it as a field operating on a Fock state, then it is difficult to see where you want to introduce the smearing. $\endgroup$ Nov 7, 2023 at 8:49
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Yes, if we are mathematically careful about the distributional nature of a quantum field, then we should indeed consistently write quantum fields only as "smeared functions" $\phi(f)$.

But really your question isn't about quantum fields, it's just about how physicists treat distributions, namely by pretending they're functions. We do this everywhere, not just in QFT: You will see classical mass densities $\rho(x) = \delta(x)$ or quantum mechanical wave "functions" $\psi(x) = \delta(x)$. The Dirac delta, too, is a distribution and not a function. See also this answer of mine for more ruminations on whether it is "wrong" to treat the $\delta$-distribution this way, and how the "wrong" language is often just shorthand for something that is also rigorously true.

For comparison, just look at how far a typical QM treatment gets by pretending that the operators on the infinite-dimensional Hilbert spaces of QM are "just like matrices", ignoring the rich subtlety of functional-analytic problems that can arise for unbounded operators. Sure, there's some edge cases where you get nonsense, but the typical intro to QM never runs into these. If you accept the usual treatment of operators in QM, you can accept the lax treatment of quantum fields as functions rather than distributions, too!

In QFT, it turns out that the distributional nature is related to something else the physicist has to do, namely UV renormalization: We use "point" interactions of the form $\phi(x)^n$ to define the interaction terms of our theories, and such a product is in general ill-defined for distributions. The physicist ignores the distributional nature of the quantum fields, pretends $\phi(x)^n$ is a meaningful thing to work with and pays for it with "infinities" we then have to "renormalize away".

The mathematically careful can instead choose to work in causal perturbation theory due to Epstein and Glaser, where "infinities" never appear and the renormalization parameters that control the UV divergences in the usual physics approach instead parametrize the choices of the products of distributions order by order in perturbation theory.

Both pathways end up with the same recipe for computing real-world scattering amplitudes depending on the renormalization parameters.

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    $\begingroup$ If I'm completely honest, this frustrates me a bit because I can't help but wonder if a considerable amount of issues we run into in physics would simply go away if more care was taken in the corresponding mathematics. I understand if both pathways end up being qualitatively the same, but, in this case at least, what is the point of working with distributions if we are going to treat them as functions anyway? Regardless, thank you for your answer, and sorry for my "rant". If I may ask you something else, do you know how smeared fields act on a Fock state? $\endgroup$
    – RMC777
    Nov 6, 2023 at 15:11
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    $\begingroup$ @RMC777 actually, physics often does better to take less care in mathematics. Formally, we often seem to sum series that we know don't converge, because we have also developed computational tools that really can subtract infinity from infinity to calculate the correct answer to an experiment to an impressive number of decimal places. In that regard, QED is the most successful scientific theory ever invented — but that is because it is sloppy with mathematics. As long as we get close enough to the right answer, the apparent ridiculousness of the procedure doesn't matter. $\endgroup$
    – Ryan C
    Nov 6, 2023 at 19:24
  • $\begingroup$ I disagree with both of the above comments about the role of rigor :) $\endgroup$
    – ACuriousMind
    Nov 6, 2023 at 20:47
  • $\begingroup$ The wrong maths is exactly the problem. $\endgroup$
    – phil soady
    Nov 8, 2023 at 10:52
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I am studying these topics and I think that an answer from who is beginning to manage this field, could be helpful because often same questions arise for different people. You should always think of an operator as a functional. What I mean, is that the most important think in QFT are the Green functions, i.e., objects of the the form

$$G(x_1,\dots,x_n)=<\Omega| \Phi(x_1)\dots\Phi(x_n)|\Omega>$$

where $\Omega$ is some particle state of interest in a QFT (often taken to be the vacuum state of the theory, if any) and $x_1,\dots,x_n$ are spacetime points. Now, the Wightman theory (I suggest you to give al look to "Local Quantum Theory" of Rudolf Haag, one of the first chapters) things like

$$G(f_1,\cdots,f_n)=<\Omega|\Phi(f_1)\cdots\Phi(f_n)|\Omega>$$

become to be relevant, with $f_1,\dots,f_n$ some smearing functions is some functional space, and

$$\Phi(f_k)=\int_M f_k(x)\Phi(x)d^4x$$

where $M$ is the Minkowski spacetime. These objects are useful for two reasons:

  1. Functions of field operators, generate the observable von Neumann algebra. Then, one focuses on von Neumann algebras generated by some special QFT (notice that $\Phi(f_k)$ is a function of $\Phi$).

  2. If one wants to obtain information about the values which an observable can take, than there is the necessity to smear out the filed operator in such a way that the support of the smearing function relates to the sensibility of the measure. So, because you don't have exact values for an observables, than you smear them out.

To finish, $\Phi(f_k)$ is an operator, so it can act on kets. Take a closer look to the standard second quantization of Klein-Gordon filed to clarify further your doubts.

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